Explain how the elimination of a predator from an ecosystem might result in starvation amongst its prey species.

A single atom of an element has 21 nutrons 20 electtons and 20 protons which element is it

guajiro [1.7K]

your element would be calcium

7 0

3 years ago

If the decomposition of a sample of k c l o 3 kclox3 produces 3. 29 g of o 2 ox2 , what was the mass (g) of the original sample?

____ [38]

The mass (g) of the original sample after decomposition is 8.3983 g.

A decomposition reaction can be described as a chemical reaction wherein one reactant breaks down into or extra merchandise.

explanation:

Reaction 2KClO₃ ⇒ 2KCl + 3O₂

moles 2 2 3

molar mass 122.55 74.55 32

Given, Mass of O₂ = 3.29g ⇒ moles of O₂

= (3.29/32) = 0.1028

3 moles of O₂ produced by 2 moles of KClO₃

Therefore, 0.1028 moles of O₂ produced by (2*0.1028/3) = 0.06853 moles of Kclo₃

Mass of KClo₃ in original sample is = moles * molar mass

= 0.06853 * 122.55

= 8.3983 g

A decomposition response occurs whilst one reactant breaks down into or extra merchandise. this may be represented through the general equation: XY → X+ Y. Examples of decomposition reactions consist of the breakdown of hydrogen peroxide to water and oxygen, and the breakdown of water to hydrogen and oxygen.

Learn more about decomposition here:-brainly.com/question/27300160

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5 0

1 year ago

Calculate the pH of 1.00 L of a buffer that contains 0.105 M HNO2 and 0.170 M NaNO2. What is the pH of the same buffer after the

kari74 [83]

Answer:

1.- pH =3.61

2.-pH =3.53

Explanation:

In the first part of this problem we can compute the pH of the buffer by making use of the Henderson-Hasselbach equation,

pH = pKa + log [A⁻]/[HA]

where [A⁻] is the conjugate base anion concentration ( [NO₂⁻]), [HA] is the weak acid concentration,[HNO₂].

In the second part, our strategy has to take into account that some of the weak base NO₂⁻ will be consumed by reaction with the very strong acid HCl. Thus, first we will calculate the new concentrations, and then find the new pH similar to the first part.

First Part

pH = 3.40+ log {0.170 /0.105}

pH = 3.61

Second Part

# mol HCl = ( 0.001 L ) x 12.0 mol / L = 0.012

# mol NaNO₂ reacted = 0.012 mol ( 1: 1 reaction)

# mol NaNO₂ initial = 0.170 mol/L x 1 L = 0.170 mol

# mol NaNO₂ remaining = (0.170 - 0.012) mol = 0.158

# mol HNO₂ produced = 0.012 mol

# mol HNO₂ initial = 0.105

# new mol HNO₂ = (0.105 + 0.012) mol = 0.117 mol

Now we are ready to use the Henderson-Hasselbach with the new ration. Notice that we dont have to calculate the concentration (M) since we are using a ratio.

pH = 3.40 + log {0.158/.0117}

pH = 3.53

Notice there is little variation in the pH of the buffer. That is the usefulness of buffers.

4 0

2 years ago

Read 2 more answers

Methanol, ethanol, and n−propanol are three common alcohols. When 1.00 g of each of these alcohols is burned in air, heat is lib

KengaRu [80]

Answer:

For methanol: Heat of combustion = -22.6 kJ / 0.0312 moles = -724.3590 kJ/mol (negative sign signifies release of heat)

For ethanol: Heat of combustion = -29.7 kJ / 0.0217 moles = -1368.6636 kJ/mol (negative sign signifies release of heat)

For propanol: Heat of combustion = -33.4 kJ / 0.0166 moles = -2012.0482 kJ/mol (negative sign signifies release of heat)

Explanation:

Given:

Mass of Methanol = 1.0 g

Mass of ethanol = 1.00 g

Mass of n-propanol = 1.00 g

For methanol:

2 CH₃OH + 3 O₂ ----> 2 CO₂ + 4 H₂O, ∆H₀ = -22.6 kJ/g (negative sign signifies release of heat)

1 g of methanol on combustion gives 22.6 kJ of energy

Calculation of moles of methanol:

$moles=\frac{Mass(m)}{Molar\ mass (M)}$

Molar mass of methanol = 32.04 g/mol

Thus moles of methanol = 1 g/ (32.04 g/mol) = 0.0312 moles

Hence energy in kJ/mol:

Heat of combustion = -22.6 kJ / 0.0312 moles = -724.3590 kJ/mol (negative sign signifies release of heat)

For ethanol:

C₂H₅OH + 3 O₂ ----> 2 CO₂ + 3 H₂O, ∆H₀ = -29.7 kJ/g (negative sign signifies release of heat)

1 g of ethanol on combustion gives 29.7 kJ of energy

Calculation of moles of ethanol:

$moles=\frac{Mass(m)}{Molar\ mass (M)}$

Molar mass of ethanol = 46.07 g/mol

Thus moles of ethanol = 1 g/ (46.07 g/mol) = 0.0217 moles

Hence energy in kJ/mol:

Heat of combustion = -29.7 kJ / 0.0217 moles = -1368.6636 kJ/mol (negative sign signifies release of heat)

For propanol:

2 C₃H₇OH + 9 O₂ ----> 6 CO₂ + 8 H₂O, ∆H₀ = -33.4 kJ/g , (negative sign signifies release of heat)

1 g of methanol on combustion gives 33.4 kJ of energy

Calculation of moles of methanol:

$moles=\frac{Mass(m)}{Molar\ mass (M)}$

Molar mass of methanol = 60.09 g/mol

Thus moles of methanol = 1 g/ (60.09 g/mol) = 0.0166 moles

Hence energy in kJ/mol:

Heat of combustion = -33.4 kJ / 0.0166 moles = -2012.0482 kJ/mol (negative sign signifies release of heat)

5 0

2 years ago

Matter is made up of particle. Which of the following statement is true about these particle? A. A partical that make up soild d

Andrews [41]

Answer: Answer A

Explanation: sorry if I'm wrong but I'm sharing what I know

8 0

3 years ago